Question

a concentration cell was made using a silver metal electrode in a saturated solution of agi for the anode half - cell, and a silver metal electrode in 1.00 m ag⁺ for the cathode half - cell. ( e_{\text{cell}} ) for the cell was 0.475 v at 25.0 °c. calculate the ( k_{sp} ) for agi. ( r = 8.3145space jcdot mol^{-1}k^{-1} ) and ( f = 96486space jcdot v^{-1}mol^{-1} ) provide an answer to three significant figures.

Explanation:

Step1: Determine the cell reaction and \(n\)

The cell reaction is \(Ag^{+}(1.00\ M)\to Ag^{+}(satd\ AgI)\). The number of electrons transferred \(n = 1\)

Step2: Use the Nernst equation

The Nernst equation for a concentration cell is \(E = E^{\circ}-\frac{RT}{nF}\ln\frac{[Ag^{+}]_{anode}}{[Ag^{+}]_{cathode}}\). Since \(E^{\circ}=0\) (same electrodes and same - type reactions), we have \(E=\frac{RT}{nF}\ln\frac{[Ag^{+}]_{cathode}}{[Ag^{+}]_{anode}}\)

Substitute \(E = 0.475\ V\), \(R=8.3145\ J\cdot mol^{-1}K^{-1}\), \(T=(25 + 273.15)K=298.15\ K\), \(n = 1\), \(F = 96486\ J\cdot V^{-1}mol^{-1}\), and \([Ag^{+}]_{cathode}=1.00\ M\) into the Nernst equation:

$$0.475=\frac{8.3145\times298.15}{1\times96486}\ln\frac{1}{[Ag^{+}]_{anode}}$$

First, calculate \(\frac{8.3145\times298.15}{96486}\approx0.02576\)

Then, \(\ln\frac{1}{[Ag^{+}]_{anode}}=\frac{0.475}{0.02576}\approx18.44\)

\(\frac{1}{[Ag^{+}]_{anode}}=e^{18.44}\approx1.16\times10^{8}\), so \([Ag^{+}]_{anode}\approx8.62\times 10^{-9}\ M\)

For \(AgI(s)
ightleftharpoons Ag^{+}(aq)+I^{-}(aq)\), in a saturated \(AgI\) solution, \([Ag^{+}]=[I^{-}]\) (from the dissolution of \(AgI\))

Step3: Calculate \(K_{sp}\)

\(K_{sp}=[Ag^{+}][I^{-}]\). Since \([Ag^{+}]=[I^{-}]\) (from \(AgI\) dissolution) and \([Ag^{+}]_{anode}\) is the concentration of \(Ag^{+}\) in the saturated \(AgI\) solution

\(K_{sp}=(8.62\times 10^{-9})^{2}\)

$$K_{sp}=7.43\times10^{-17}$$

Answer:

\(7.43\times 10^{-17}\)